Ch7 Statistics Dr. Deshi Ye [email protected]. 2/139 Outline State what is estimated Point estimation Interval estimation Compute sample size Tests of

Ch7 Statistics

Dr. Deshi Ye

[email protected]

2/139

OutlineState what is estimated

Point estimation

Interval estimation

Compute sample size

Tests of Hypotheses

Null Hypothesis and Tests of Hypotheses

Hypotheses concern one mean

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Why Statistics?The purpose of most statistical investigations is to generalize from information contained in random samples about the populations from which the samples were obtained.

How: estimation and tests of Hypothesis

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ThinkingSuppose you’re interested in the average amount of money that students in this class (the population) have on them. How would you find out?

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Introduction to Estimation

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Statistical Methods

StatisticalMethods

DescriptiveStatistics

InferentialStatistics

EstimationHypothesis

Testing

StatisticalMethods




Testing

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Estimation Process

Mean, , is unknown

PopulationPopulation Random SampleRandom SampleI am 95%

confident that is between

40 & 60.

Mean X= 50

Sample

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Estimation Methods

Estimation

PointEstimation

IntervalEstimation

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Point estimationWe want to know the mean of a population. However, it is unavailable.

Hence, we choose a sample data, and calculate from the choosing sample data. Then estimate that the population is also the same mean.

Point estimation: concerns the choosing of a statistic, that is, a single number calculated from sample data for which we have some expectation, or assurance, that is reasonably close the parameter it is supposed to estimate.

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Point Estimation1. Provides Single Value

Based on Observations from 1 Sample

2. Gives No Information about How Close Value Is to the Unknown Population Parameter

3. Example: Sample MeanX = 3 Is Point Estimate of Unknown Population Mean

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Point estimation of a meanParameter: Population mean

Data: A random sample

Estimator:

Estimate of standard error: S

n

X

1 2, , , nX X X

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EXScientists need to be able to detect small amounts of contaminants in the environment. Sample data is listed as follows:

2.4 2.9 2.7 2.6 2.9 2.0 2.8 2.2 2.4 2.4 2.0 2.5

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EX of point estimationCompute the point estimator and estimate its standard deviation (also called the estimated standard error of ).

Solution:

X

X

29.82.483

12x

2 2 22

2 1 1

( )75.08 (29.8) /12

0.097881 1 12 1

n n

i ii i

x x x nxs

n n

n

Hence the estimated standard deviation is 0.09788 /12 0.090s

n

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Bias or UnbiasQuestion: is the estimation good enough?

EX.In previous ex, as a check on the current capabilities, the measurements were made on test specimens spiked with a known concentration 1.25 ug/l of lead. That is the readings should average 1.25 if there is no background lead in the samples. There appears to be either a bias due to laboratory procedure or some lead already in the samples before they were spiked.

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Unbiased estimatorLet be the parameter of interest and

be a statistic.

A statistic is said to be an unbiased estimator, or its value an unbiased estimate, if and only if the mean of the sampling distribution of the estimator equals , whatever the value of .

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More efficient unbiased estimatorEstimator is not unique: for example: it can be shown that for a random sample of size n=2, the mean as well as the weighted mean where a, b are positive constants, are unbiased estimates of the mean of the population.

A statistics is said to be a more efficient unbiased estimator of the parameter than the statistics if

1. and are both unbiased estimators of 2. the variance of the sampling distribution of the first

estimator is no larger than that of the second and is smaller for at least one value of .

1 2

12

1 2

2

X X

1 2aX bX

a b

Finding Sample Sizes

I don’t want to sample too much or too little!

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The error The error between the estimator and the quantity it is supposed to estimate is: X

/

X

n

is a random variable having approximately the standard normal distribution

We could assert with probability that the inequality

/ 2 / 2/

Xz z

n

1

/ 2( ) 1 / 2P Z z Remind that

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Maximum error of estimateThe error will be less than

with probability .1

| |X

/ 2E zn

Specially, / 2 1.96, 0.05z when

/ 2 2.575, 0.01z when

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EXAn industrial engineer intends to use the mean of a random sample of size n=150 to estimate the average mechanical aptitude of assembly line workers in a large industry. If, on the basis of experience, the engineer can assume that for such data, what can he assert with probability 0.99 about the maximum size of his error.

6.2

6.22.575 1.3

150E

Thus, the engineer can assert with probability 0.99 that his error is at most 1.3

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Determine of sample sizeSuppose that we want to use the mean of a large random sample to estimate the mean of a population, and want to be able to assert with probability that the error will be at most some prescribed quantity E. As before, we get

2/ 2( )z

nE

1

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EXA research worker want to determine the average time it takes a mechanic to rotate the tires of a car, and she wants to be able to assert with 95% confidence that the mean of her sample is off by at most 0.5 minute. If she can presume from past experience that minutes, how large a sample will she have to take?

1.6

21.96 1.6( ) 39.3

0.5n

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ContinuousThe method discussed so far requires be known or it can be approximated with the sample standard deviation s, thus requiring that n be large. Another approach: if it is reasonable to assume that we are sampling from a normal population, we get

is a random variable having the t distribution with n-1 degree of freedom.

/

Xt

S n

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EXThe first example, n=12, 0.09788 /12 0.090

s

n

0.01 2.718t For n=11 degrees of freedom

0.01 2.718 0.09 0.2s

tn

2.483 1.25 1.233 0.2X

Thus, one can assert with 98% confidence that the maximum error is within 0.2

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Estimation Methods

Estimation

PointEstimation

IntervalEstimation

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Interval Estimation

1. Provides Range of Values Based on Observations from 1 Sample

2. Gives Information about Closeness to Unknown Population Parameter

Stated in terms of Probability

3. Example: Unknown Population Mean Lies Between 50 & 70 with 95% Confidence

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Interval Estimation

Sample statistic Sample statistic

(point estimate)(point estimate)

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Interval Estimation

Confidence Confidence intervalinterval



Confidence Confidence limit (lower)limit (lower)

Confidence Confidence limit (upper)limit (upper)

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Interval Estimation

Confidence Confidence intervalinterval



Confidence Confidence limit (lower)limit (lower)

Confidence Confidence limit (upper)limit (upper)

A A probabilityprobability that the population parameter that the population parameter falls somewhere within the interval.falls somewhere within the interval.

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Confidence Interval Estimates

ConfidenceIntervals

ProportionMean Variance

UnknownKnown

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Confidence Interval Mean ( Known)

AssumptionsPopulation standard deviation is known

Population is normally distributed

If not normal, can be approximated by normal distribution (n 30)

Confidence Interval Estimate

/ 2 / 2X Z X Zn n

/ 2 / 2X Z X Zn n

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Interval EstimationInterval estimation: with intervals for which we can assert with a reasonable degree of certainty that they will contain the parameter under consideration.

For a large random sample (n > 30) from a population with the unknown mean and the known variance. When the observed value becomes available, we obtainx

/ 2 / 2x z x zn n

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Confidence intervalWe can claim with confidence that the interval

(1 )100%

/ 2 / 2[ , ]x z x zn n

Contains

It is customary to refer to an interval of this kind as a confidence interval for having the degree of confidence

(1 )100%

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Confidence Interval Estimates

ConfidenceIntervals

ProportionMean Variance

UnknownKnown

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Solution for Small Samples1. Assumptions

Population of X Is Normally Distributed

Use Student’s t Distribution1.Define variable

2.T has the Student distribution with n -1 degrees of freedom (When X is normally distributed)

• There’s a different Student distribution for different degrees of freedom

• As n gets large, Student distribution approximates a normal distribution with mean = 0 and sigma = 1

/

XT

s n

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Small sample (n<30)Small sample and we assume to get sampling from a normal distribution population.

We get the confidence interval formula

(1 )100%

/ 2 / 2

s sx t x t

n n

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EX.The mean weight loss of n=16 grinding balls after a certain length of time in mill slurry is 3.42 grams with a standard deviation of 0.68 gram. Construct a 99% confidence interval for the true mean weight loss of such grinding balls under the stated condition.

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Confidence Depends on Interval (z)

90% Samples90% Samples



+1.65+1.65x x +2.58+2.58xx

xx__

XX

+1.96+1.96xx

-2.58-2.58xx -1.65-1.65xx

-1.96-1.96xx

XX= = ± Z ± Zxx

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1. Probability that the Unknown Population Parameter Falls Within Interval

2. Denoted (1 - Is Probability That Parameter Is Not Within Interval

3. Typical Values Are 99%, 95%, 90%

Confidence Level

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Intervals & Confidence Level

x =

1 - /2/2

X_

x_

x =

1 - /2/2

X_

x_Sampling Sampling

Distribution Distribution of Meanof Mean

Intervals derived from Intervals derived from many samplesmany samples

Intervals Intervals extend from extend from X - ZX - ZXX to to

X + ZX + ZXX

(1 - (1 - ) % of ) % of intervals intervals contain contain . .

% do not.% do not.

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Confidence interval & levelIt is useful to think confidence intervals as a range of "plausible" values for the parameter. confidence interval is different from interpreting the confidence level

suppose we've taken a random sample of 10 ice-cream cones, and determined that a 95% confidence interval for the mean caloric contents of a single scoop of ice-cream is (260,310). Interpret the confidence level: If we repeatedly took samples of size 10 and then formed confidence intervals, we would expect 95% of them to contain the true (but unknown) mean. Interpret this particular confidence interval: we are 95% confident that the true mean caloric content lies between 260 and 310.

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Confidence interval & levelThe wider the confidence interval you are willing to accept, the more certain you can be that the whole population answers would be within that range. For example, if you asked a sample of 1000 people in a city which brand of cola they preferred, and 60% said Brand A, you can be very certain that between 40 and 80% of all the people in the city actually do prefer that brand, but you cannot be so sure that between 59 and 61% of the people in the city prefer the brand.

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1. Data DispersionMeasured by

2. Sample Size

3. Level of Confidence (1 - )

Affects Z

Factors Affecting Interval Width

Intervals Extend fromIntervals Extend from

X - ZX - ZXX to toX + ZX + ZXX

© 1984-1994 T/Maker Co.

/x n

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Statistical Methods

StatisticalMethods




Testing

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Hypothesis Testing

PopulationPopulation

I believe the population mean age is 50 (hypothesis).


Reject hypothesis! Not close.


MeanMean

X X = 20= 20

Random Random samplesample

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What’s a Hypothesis?

A belief about a population parameter

Parameter is Population mean, proportion, variance

Must be statedbefore analysis

I believe the mean GPA I believe the mean GPA of this class is 3.8!of this class is 3.8!

© 1984-1994 T/Maker Co.

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Tests of HypothesesSuppose that a consumer protection agency wants to test a paint manufacturer’s claim that the average drying time of his new “fast-drying” paint is 20 minutes. It instructs a member of its research staff to paint each of 36 boards using a different 1-gallon can of the paint, with the intention of rejecting the claim if the mean of the drying time exceeds 20.75 minutes. Otherwise, it will accept the claim.Question: Is it a infallible criterion for accepting or rejecting the claim?

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EX cont.Assuming that it is known from past experience that the

standard deviation 2.4

Let us investigate the possibility that the sample may exceed 20.75 minutes even though the true mean is 20 minutes

20 20.75

x Minutes

Reject the claim

that 20 Accept the claim

that 20

0.0304

20.75 201.875

2.4 / 36z

The probability of erroneously rejecting the hypothesis that 20

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Another possibilityThe procedure fails to detect that 20

21 20.75

0.2660

20

Accept the claim

that

Reject

20

the claim

that

20.75 210.625

2.4 / 36z

Suppose that the true mean of drying time is 21

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Type of errorsH: the hypothesis. Ex.

Accept H Reject H

H is true Correct decision Type I error

H is false Type II error Correct decision

20

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Type I error: If hypotheses is true but rejected. Denoted by the letter

EX.

Type II error: If hypotheses is false but not rejected. Denoted by the letter

EX.

0.0304

0.2660

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7.4 Null HypothesesQuestion: Can we formulate minutes, where can take on more than one possible value?

Null Hypotheses : (Pronounced H-nought) Usually require that we hypothesize the opposite of what we hope to prove.

EX. If we want to show that one method of teaching computer programming is more efficient than another, we hypothesize that the two methods are equally effective.

The null hypothesis proposes something initially presumed true. It is rejected only when it becomes evidently false.

20

0H

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Null Hypothesis

1. What is tested

2. has serious outcome if incorrect decision made

3. Designated H0

4. Specified as H0: Some Numeric Value

Specified with = Sign Even if , or Example, H0: 3

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Alternative Hypothesis

1. Opposite of Null Hypothesis

2. Always Has Inequality Sign: ,, or

3. Designated Ha

4. Specified Ha: < Some Value

Example, Ha: < 3

will lead to two-sided tests

<, > will lead to one-sided tests

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AlternativeOne-sided alternative: In the drying time example, the null hypothesis is minutes and the alternative hypothesis is

Two-sided alternative: where is the value assumed under the null hypothesis.

20

20

0 0

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Selecting the null hypothesis Guideline for selecting the null hypothesis: When the goal of an experiment is to establish an assertion, the negation of the assertion should be taken as the null hypothesis. The assertion becomes the alternative hypothesis.

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Identifying Hypotheses Steps

1. Example Problem: Test That the Population Mean Is Not 3

2. StepsState the Question Statistically ( 3)

State the Opposite Statistically ( = 3)Must Be Mutually Exclusive & Exhaustive

Select the Alternative Hypothesis ( 3)Has the , <, or > Sign

State the Null Hypothesis ( = 3)

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State the question statistically: = 12

State the opposite statistically: 12

Select the alternative hypothesis: Ha: 12

State the null hypothesis: H0: = 12

Is the population average amount of TV Is the population average amount of TV viewing 12 hours?viewing 12 hours?

What Are the Hypotheses?

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State the question statistically: 12

State the opposite statistically: = 12


State the null hypothesis: H0: = 12

Is the population average amount of TV Is the population average amount of TV viewing viewing differentdifferent from 12 hours? from 12 hours?


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State the null hypothesis: H0: 20

Is the average cost per hat less than or Is the average cost per hat less than or equal to $20?equal to $20?


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State the null hypothesis: H0: 25

Is the average amount spent in the Is the average amount spent in the bookstore greater than $25?bookstore greater than $25?


Decision Making Risks

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Errors in making decision1. Type I Error

Reject True Null HypothesisHas Serious ConsequencesProbability of Type I Error Is (Alpha)

Called Level of Significance

2. Type II ErrorDo Not Reject False Null HypothesisProbability of Type II Error Is (Beta)

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Jury Trial H0 Test

Actual Situation Actual Situation

Verdict Innocent Guilty Decision H0 True H0 False

Innocent Correct Error Do Not Reject

H0 1 - Type II

Error ()

Guilty Error Correct Reject H0

Type I Error ()

Power (1 - )

Jury Trial H0 Test



Innocent Correct Error Do Not Reject

H0 1 - Type II

Error ()


Type I Error ()

Power (1 - )

Decision ResultsHH00: Innocent: Innocent

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Jury Trial H0 Test



Innocent Correct Error

Accept H0

1 - Type II Error

()


Type I Error ()

Power (1 - )

Jury Trial H0 Test



Innocent Correct Error

Accept H0

1 - Type II Error

()


Type I Error ()

Power (1 - )

Decision ResultsHH00: Innocent: Innocent

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& have an inverse relationship

You can’t reduce both errors simultaneously!

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Factors Affecting 1. True Value of Population Parameter

Increases When Difference With Hypothesized Parameter Decreases

2. Significance Level, Increases When Decreases

3. Population Standard Deviation, Increases When Increases

4. Sample Size, n

Increases When n Decreases

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Hypothesis Testing

PopulationPopulation





MeanMean

X X = 20= 20

Random Random samplesample

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Basic Idea

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Basic Idea

Sample Mean = 50 Sample Mean = 50

HH00HH00

Sampling DistributionSampling Distribution

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Basic Idea



It is unlikely It is unlikely that we would that we would get a sample get a sample mean of this mean of this value ...value ...

20202020HH00HH00

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Basic Idea




... if in fact this were... if in fact this were the population mean the population mean

20202020HH00HH00

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Basic Idea




... if in fact this were... if in fact this were the population mean the population mean

... therefore, ... therefore, we reject the we reject the hypothesis hypothesis

that that = 50.= 50.

20202020HH00HH00

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Hypothesis testing1. We formulate a null hypothesis and an appropriate alternative hypothesis which we accept when the null hypothesis must be rejected.

2. We specify the probability of a Type I error. If possible, desired, or necessary, we may also specify the probabilities of Type II errors for particular alternatives.

3. Based on the sampling distribution of an appropriate statistic, we construct a criterion for testing the null hypothesis against the given alternative.

4. We calculate from the data the value of the statistic on which the decision is to be based.

5. We decide whether to reject the null hypothesis or whether to fail to reject it.

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Level of Significance1. Probability

2. Defines Unlikely Values of Sample Statistic if Null Hypothesis Is True

Called Rejection Region of Sampling Distribution

3. Designated (alpha)

4. Selected by Researcher at Start

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Level of significanceThe probability of a Type I errorType I error is also called the level of significance. Usually, we set .

Step 2 can often be performed even when the null hypothesis specifies a range of values for the parameter.

Ex. The null hypothesis:Then we can claim that

0.05 0.01or

20

0.0304

In general, we can only specify the maximum probability of a Type I error, and by again the worst possibility.

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Criterion One-sided criterion (one-sided test or one-tailed test): ex. One-sided alternative

Two-sided criterion (two-sided test or two-tailed test): ex. two-sided alternative

20

4

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Rejection Region (One-Tail Test)

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1 - 1 -

Level of ConfidenceLevel of Confidence


H0

valueCriticalvalue

Sample Statistic

RejectionRegion

NonrejectionRegion

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HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion


1 - 1 -


Observed sample statisticObserved sample statistic


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HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion


1 - 1 -



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Rejection Region (Two-Tailed Test)

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1 - 1 -



HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion

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1 - 1 -



HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


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1 - 1 -



HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


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1 - 1 -



HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


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H0 Testing Steps

Set up critical values

Collect data

Compute test statistic

Make statistical decision

Express decision

State HState H00

State HState Haa

Choose Choose

Choose Choose nn

Choose testChoose test

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One Population Tests

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

Mean Proportion Variance

2 Test(1 & 2tail)

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7.5 Hypothesis concerning one mean

Suppose we want to test on the basis of n=35 determinations and at the 0.05 level of significance whether the thermal conductivity of a certain kind of cement brick is 0.340, as has been claimed. We can expect that the variability of such determinations is given by 0.010

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Solution1) Null hypothesis:

alternative hypothesis:

2) Level of significance:

3) Criterion:

0.340

/

XZ

n

z

0.05

0.340

/ 2z/ 2z

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Criterion Region for testing (Normal population and known)

Alternative hypothesis

Reject null hypothesis if

0

0

0

0

Z z

Z z

/ 2 / 2Z z or Z z

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Critical values

One-sided alternatives

Two-sided alternatives

-1.645

1.645

-1.96

1.96

-2.33

2.33

-2.575

2.575

0.05

0.01

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EX. Cont.3) Criterion: Reject the null hypothesis if

1.96 1.96Z or Z

/

XZ

n

where

4) Calculations:0.343 0.34

1.770.01/ 35

z

5) Decision: The null hypothesis cannot be rejected.

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P-valueP value (or tail probability): the probability of getting difference between and

greater than or equal to that actually observed.

EX. In above example

1.771.77

( )P Z z

x 0

2(1 0.9616) 0.078

½ P-value½ P-value

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P-Value1. Probability of Obtaining a Test Statistic More Extreme (or than Actual Sample Value Given H0 Is True

2. Called Observed Level of SignificanceSmallest Value of H0 Can Be Rejected

3. Used to Make Rejection DecisionIf p-Value , Do Not Reject H0

If p-Value < , Reject H0

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P valueP value for a given test statistic and null hypothesis: The P value is the probability of obtaining a value for the test statistic that is as extreme or more extreme than the value actually observed. Probability is calculated under the null hypothesis.

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EXA process for producing vinyl floor covering has been stable for a long period of time, and the surface hardness measurement of the flooring produced has a normal distribution with mean 4.5 and standard deviation 1.5. A second shift has been hired and trained and their production needs to be monitored. Consider testing the hypothesis

0 1: 4.5 : 4.5H versus H

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A random sample of hardness measurements is made of n=25 vinyl specimens produced by the second shift. Calculate the P value when using the test statistic

If

/

XZ

n

3.9X

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SolutionThe observed value of the test statistic is

3.9 4.52.0

1.5 / 25z

Since the alternative hypothesis is two-sided, we must consider large negative value for Z as well as large positive values

( 2.0) ( 2.0) 0.0228P Z P Z

0.0456Consequently, the P value is

0.05 The small P value suggests the mean of the second shift is not at the target value of 4.5

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P valueTo understand P values, you have to understand fixed level testing.

With fixed level testing, a null hypothesis is proposed (usually, specifying no treatment effect) along with a level for the test, usually 0.05. All possible outcomes of the experiment are listed in order to identify extreme outcomes that would occur less than 5% of the time in aggregate if the null hypothesis were true.

This set of values is known as the critical region. They are critical because if any of them are observed, something extreme has occurred. Data are now collected and if any one of those extreme outcomes occur the results are said to be significant at the 0.05 level. The null hypothesis is rejected at the 0.05 level of significance and one star (*) is printed somewhere in a table. Some investigators note extreme outcomes that would occur less than 1% of the time and print two stars (**) if any of those are observed.

Many researchers quickly realized the limitations of reporting only whether a result achieved the 0.05 level of significance. Was a result just barely significant or wildly so? Would data that were significant at the 0.05 level be significant at the 0.01 level? At the 0.001 level? Even if the result are wildly statistically significant, is the effect large enough to be of any practical importance?

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P valueObserved significance level (or P value)--the smallest fixed level at which the null hypothesis can be rejected. If your personal fixed level is greater than or equal to the P value, you would reject the null hypothesis.

If your personal fixed level is less than to the P value, you would fail to reject the null hypothesis.

For example, if a P value is 0.027, the results are significant for all fixed levels greater than 0.027 (such as 0.05) and not significant for all fixed levels less than 0.027 (such as 0.01). A person who uses the 0.05 level would reject the null hypothesis while a person who uses the 0.01 level would fail to reject it.

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P-Value Thinking Challenge

You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. What is the value of the observed level of significance (p-Value)?

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p-Value Solution*

Z0-2.65

p-Value.004

Z0-2.65

p-Value.004

Z value of Z value of sample statisticsample statistic

From Z table: From Z table: lookup 2.65lookup 2.65

.4960.4960

Use Use alternative alternative hypothesis hypothesis to find to find directiondirection

.5000.5000-- .4960.4960

.0040.0040

p-Value is P(Z p-Value is P(Z -2.65) = .004. -2.65) = .004.p-Value < (p-Value < ( = .01). Reject H = .01). Reject H00..

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One Population Tests

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

Mean Proportion Variance

2 Test(1 & 2tail)

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t Test for Mean ( Unknown)1. Assumptions

Population Is Normally Distributed

If Not Normal, Only Slightly Skewed & Large Sample (n 30) Taken

2. Parametric Test Procedure

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Statistic for small sample

/

XT

S n

The test of null hypothesis on the statistic

0

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Criterion Region for testing (Statistic for small sample )

Alternative hypothesis

Reject null hypothesis if

0

0

0

0

T t

T t

/ 2 / 2T t or T t

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t0 t0

Two-Tailed t TestFinding Critical t Values

Given: n = 3; Given: n = 3; = .10 = .10

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t0 t0

/2 = .05/2 = .05

/2 = .05/2 = .05

Given: n = 3; Given: n = 3; = .10 = .10


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t0 t0

/2 = .05/2 = .05

/2 = .05/2 = .05

Given: n = 3; Given: n = 3; = .10 = .10

df = n - 1 = 2df = n - 1 = 2


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v t.10 t.05 t.025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182

v t.10 t.05 t.025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182t0 t0

Critical Values of t Table Critical Values of t Table (Portion)(Portion)

/2 = /2 = .05.05

/2 = .05/2 = .05

Given: n = 3; Given: n = 3; = .10 = .10

df = n - 1 = df = n - 1 = 22


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v t.10 t.05 t.025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182

v t.10 t.05 t.025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182t0 2.920-2.920 t0 2.920-2.920

Critical Values of t Table Critical Values of t Table (Portion)(Portion)

/2 = .05/2 = .05

/2 = .05/2 = .05

Given: n = 3; Given: n = 3; = .10 = .10

df = n - 1 = 2df = n - 1 = 2


113/139

One-Tailed t TestYou’re a marketing analyst

for Wal-Mart. Wal-Mart had teddy bears on sale last week. The weekly sales ($ 00) of bears sold in 10 stores was: 8 11 0 4 7 8 10 5 8 3. At the .05 level, is there evidence that the average bear sales per store is more than 5 ($ 00)?

114/139

One-Tailed t Test Solution*

H0:

Ha:

=

df =

Critical Value(s):

Test Statistic: Test Statistic:

Decision:Decision:

Conclusion:Conclusion:

115/139


H0: = 5

Ha: > 5

=

df =

Critical Value(s):


Decision:Decision:


116/139


H0: = 5

Ha: > 5

= .05

df = 10 - 1 = 9

Critical Value(s):


Decision:Decision:


117/139t0 1.8331

.05

Reject

t0 1.8331

.05

Reject


H0: = 5

Ha: > 5

= .05

df = 10 - 1 = 9

Critical Value(s):


Decision:Decision:


118/139t0 1.8331

.05

Reject

t0 1.8331

.05

Reject


H0: = 5

Ha: > 5

= .05

df = 10 - 1 = 9

Critical Value(s):


Decision:Decision:


tX

Sn

6 4 5

3 37310

131..

.tX

Sn

6 4 5

3 37310

131..

.

119/139t0 1.8331

.05

Reject

t0 1.8331

.05

Reject


H0: = 5

Ha: > 5

= .05

df = 10 - 1 = 9

Critical Value(s):


Decision:Decision:


tX

Sn

6 4 5

3 37310

131..

.tX

Sn

6 4 5

3 37310

131..

.

Do not reject at Do not reject at = .05 = .05

120/139t0 1.8331

.05

Reject

t0 1.8331

.05

Reject


H0: = 5

Ha: > 5

= .05

df = 10 - 1 = 9

Critical Value(s):


Decision:Decision:


tX

Sn

6 4 5

3 37310

131..

.tX

Sn

6 4 5

3 37310

131..

.

Do not reject at Do not reject at = .05 = .05

There is no evidence There is no evidence average is more than 5average is more than 5

Documents

Ch7 Statistics Dr. Deshi Ye [email protected]. 2/139 Outline State what is estimated Point estimation Interval estimation Compute sample size Tests of